Oulipo Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 35893 Accepted: 14488
Description
The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book:
Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…
Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive 'T's is not unusual. And they never use spaces.
So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {'A', 'B', 'C', …, 'Z'} and two finite strings over that alphabet, a word W and a text T, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap.
Input
The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:
One line with the word W, a string over {'A', 'B', 'C', …, 'Z'}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W).One line with the text T, a string over {'A', 'B', 'C', …, 'Z'}, with |W| ≤ |T| ≤ 1,000,000.Output
For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T.
Sample Input
3 BAPC BAPC AZA AZAZAZA VERDI AVERDXIVYERDIANSample Output
1 3 0Source
BAPC 2006 Qualification刚学的kmp，虽然是基础题目，但也是学了1天的成果，有详细的说明
<span style="font-size:18px;">#include<cstdio> #include<cstring> char w[10000+11]; char t[1000000+11]; int next[10000+11]; void get_next() { next[1]=0; int i=2,j,lenw=strlen(w);//i=2因为后缀最少是2，相对应的前缀是1 for(j=0;i<lenw;++i) { while(j>0&&w[i]!=w[j+1]) j=next[j]; if(w[i]==w[j+1]) ++j; next[i]=j; //这里包含2中情况j=0；如果匹配，那next[i]=j+1=1;如果不匹配代表着w[i]和第一个都不匹配，直接 //next[i]=0;执行i++;知道找到与 j+1=1匹配的i值，当然前面不匹配的就直接赋值为0 ；j!=0那就直接next[i]=j //因为在while循环里已经找到了匹配的情况 } } int times_kmp() { int ans=0,i,j; int lenw=strlen(w),lent=strlen(t); for(i=1,j=0;i<lent;++i) { while(j>0&&t[i]!=w[j+1]) j=next[j]; if(t[i]==w[j+1]) ++j; if(j==lenw-1) { ++ans; j=next[j];//其实到这里已经匹配完了，这时候的模拟串已经完了，这时候(你就假设后面还有 //并且和目标串i+1肯定不匹配，因为模拟串都不知道是谁，你怎么匹配，那就回溯吧) } } return ans; } int main() { int text; scanf("%d",&text); getchar(); while(text--) { int i=1,j=1; char ch; memset(w,'\0',sizeof(w));//这个是必须要处理的，因为上次的数据没办法清除 memset(t,'\0',sizeof(t)); while((ch=getchar())!='\n') w[i++]=ch; while((ch=getchar())!='\n') t[j++]=ch; w[0]='*';t[0]='*';//这个是什么都行，但不要是'\0'; get_next(); int ans=times_kmp(); printf("%d\n",ans); } return 0; }</span>
